Answer: Either
(i) The asymptotes are the vertical lines \(x=1\) and \(x=3\), and the horizontal asymptote is \(y=1\).
(ii) The curve meets the horizontal asymptote at \(x=\frac{2a-3}{a-2}\).
(iii) The stationary points have \(x\)-coordinates satisfying \((a-2)x^{2}+(6-4a)x+(5a-6)=0\). Hence the curve has stationary points when \(1\lt a\lt 3\), with \(a\neq 2\) already given.
(iv) For \(a\gt 3\), the graph has two vertical asymptotes at \(x=1\) and \(x=3\), and a horizontal asymptote \(y=1\). For \(2\lt a\lt 3\), the same asymptotes occur, with the middle branch having a maximum between \(y=0\) and \(y=1\), and the outside branches having a minimum point.
Or
(i) \(S_{n+4}-5S_{n+2}+2S_{n+1}-S_n=0\).
(ii) \(S_2=10\) and \(S_4=54\).
(iii) \(S_3=-6\) and \(S_6=292\).
(iv) The required value is \(248\).
Either
Let \(y=\frac{(x-2)(x-a)}{(x-1)(x-3)}\).
(i) The denominator is zero when \(x=1\) or \(x=3\), so these are vertical asymptotes. Since the numerator and denominator both have degree 2 with leading coefficient 1, the horizontal asymptote is \(y=1\).
(ii) To find where the curve meets the horizontal asymptote, set \(y=1\):
\(\frac{(x-2)(x-a)}{(x-1)(x-3)}=1\).
So
\((x-2)(x-a)=(x-1)(x-3)\).
Expanding gives
\(x^{2}-(a+2)x+2a=x^{2}-4x+3\).
Hence
\((4-a)x=3-2a\),
so
\(x=\frac{2a-3}{a-2}\).
Therefore the curve meets its horizontal asymptote at \(x=\frac{2a-3}{a-2}\).
(iii) Write \(y=\frac{N}{D}\), where \(N=(x-2)(x-a)\) and \(D=(x-1)(x-3)\). Then
\(y'=\frac{N'D-ND'}{D^{2}}\).
Stationary points satisfy \(y'=0\), so
\(N'D-ND'=0\).
Now \(N= x^{2}-(a+2)x+2a\), so \(N'=2x-(a+2)\), and \(D=x^{2}-4x+3\), so \(D'=2x-4\).
Thus
\((2x-a-2)(x^{2}-4x+3)-(x^{2}-(a+2)x+2a)(2x-4)=0\).
Expanding and simplifying gives
\((a-2)x^{2}+(6-4a)x+(5a-6)=0\).
For stationary points to exist, this quadratic in \(x\) must have real roots, so its discriminant must be non-negative:
\((6-4a)^{2}-4(a-2)(5a-6)\ge 0\).
This simplifies to
\(a^{2}-4a+3\le 0\),
so
\((a-1)(a-3)\le 0\).
Hence \(1\le a\le 3\). Since \(a\neq 2\) is given, the set of values is \(1\le a\lt 2\) or \(2\lt a\le 3\).
(iv) The graph always has vertical asymptotes \(x=1\) and \(x=3\), and horizontal asymptote \(y=1\). The middle branch lies between the vertical asymptotes, and the outer branches lie to the left of \(x=1\) and to the right of \(x=3\).
For \(a\gt 3\), the curve does not have stationary points. The sketch therefore shows the three branches approaching the asymptotes, with the middle branch passing through the horizontal asymptote once.
For \(2\lt a\lt 3\), the curve has stationary points. The middle branch has a maximum with value between \(0\) and \(1\), and each outer branch has a minimum point.
Or
Let the roots of \(x^{4}-5x^{2}+2x-1=0\) be \(\alpha,\beta,\gamma,\delta\), and define \(S_n=\alpha^n+\beta^n+\gamma^n+\delta^n\).
(i) If \(r\) is any root, then
\(r^{4}-5r^{2}+2r-1=0\).
Multiplying by \(r^n\) gives
\(r^{n+4}-5r^{n+2}+2r^{n+1}-r^n=0\).
Applying this to each root and adding,
\(S_{n+4}-5S_{n+2}+2S_{n+1}-S_n=0\).
(ii) By Vieta's formulae, the sum of the roots is zero, so \(S_1=0\), and the sum of pairwise products is \(-5\).
Now
\(S_2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\cdots)=0^2-2(-5)=10\).
Using the recurrence with \(n=0\),
\(S_4-5S_2+2S_1-S_0=0\).
Since \(S_0=4\), this gives
\(S_4=5(10)-2(0)+4=54\).
(iii) Use the recurrence with \(n=-1\):
\(S_3-5S_1+2S_0-S_{-1}=0\).
To find \(S_{-1}\), let \(y=1/x\). Then the roots \(1/\alpha,1/\beta,1/\gamma,1/\delta\) satisfy
\(1-5y^2+2y^3-y^4=0\), or equivalently \(y^4-2y^3+5y^2-1=0\).
Its roots sum to \(2\), so \(S_{-1}=2\). Therefore
\(S_3-5(0)+2(4)-2=0\),
so \(S_3=-6\).
Now use the recurrence with \(n=2\):
\(S_6-5S_4+2S_3-S_2=0\).
Hence
\(S_6=5(54)-2(-6)+10=292\).
(iv) Consider
\(\alpha^2(\beta^4+\gamma^4+\delta^4)+\beta^2(\gamma^4+\delta^4+\alpha^4)+\gamma^2(\delta^4+\alpha^4+\beta^4)+\delta^2(\alpha^4+\beta^4+\gamma^4)\).
When expanded, this is the sum of all terms \(r^2s^4\) with distinct roots \(r\neq s\). So it equals
\((\alpha^2+\beta^2+\gamma^2+\delta^2)(\alpha^4+\beta^4+\gamma^4+\delta^4)-\left(\alpha^6+\beta^6+\gamma^6+\delta^6\right)\).
That is, \(S_2S_4-S_6\).
Therefore the required value is
\(10\cdot 54-292=248\).
